Tutoial - Part 1 TutoRial - Part 1

Marine Ecosystem Dynamics - 2024

New script

As seen during the presentation, we will keep track of our progress. We thus need to open a new script following one of the option below.

flowchart LR
  A[1. File] --> B[2. New File]
  B --> C[3. R script]

flowchart LR
  A["⌘/Ctlr + ⇧ + N"]

syntax R syntax

is a programming language that use a simplified syntax. In this section, we will explore how to write a script and execute it. R is a programming language that use a simplified syntax. In this section, we will explore how to write a script and execute it.

But first some syntax information:

• Everything after # is considered as a comment and will not be executed. It is very important to write what we are doing, so we do not get lost next time we open our scripts.

# 2 + 2 will not work because of the #
2 + 2 # We should then annotate our script like this
#> [1] 4

• Several lines of code can be written in one line but must be separated by a semicolon

2 + 2
#> [1] 4
3 * 2
#> [1] 6

# This can also be written as follow:
2 + 2 ; 3 * 2
#> [1] 4
#> [1] 6

• In we can name any object using =, <-, -> or assign In R we can name any object using =, <-, -> or assign

c(1, 2, 3, 4) -> my_first_vector
my_vector <- c(1, 2, 3, 4)
my_function = function(x){x + 2}
assign("x", c(2, 3, 4, 5))

• == is a logical function that can be translated as is equal to, contrarily is not equal to is written !=

2 + 2 == 4
#> [1] TRUE
3 * 2 == 4
#> [1] FALSE
3 * 2 != 4
#> [1] TRUE

Exercises

Using a new script, do these calculations: Using a new R script, do these calculations:

• \ 2^7

Solution

2^7
#> [1] 128

• \ cos(\pi)

Hint

?cos()
?pi()

Solution

cos(pi)
#> [1] -1

• The sum of all number from 1 to 100

Hint

Operations can take place for an entire vector

Solution

vector <- seq(from = 1, to = 100, by = 1) # Create a vector from 1 to 100
sum(vector) # Calculate the sum
#> [1] 5050

Create a parameter x1 that equals to 5 and a parameter x2 that equals to 10

Solution

x1 <- 5 ; x2 <-  10

• Is \ 2* x1 equal to x2?

Solution

2 * x1 == x2
#> [1] TRUE

Functions

As seen during the lecture, works with functions that can: As seen during the lecture, R works with functions that can:

• Already be implemented in base Already be implemented in base R
• Comming from another package
• Created by the user

We will see these three examples in this section, but first it is important to remember that the typical structure of a function is function(argument1, ...).

Fortunately helps us to remember what are the needed arguments: Fortunately R helps us to remember what are the needed arguments:

• Using help() or ?

help(topic = "sin")
?sin

• Using example

example(sum)
#> 
#> sum> ## Pass a vector to sum, and it will add the elements together.
#> sum> sum(1:5)
#> [1] 15
#> 
#> sum> ## Pass several numbers to sum, and it also adds the elements.
#> sum> sum(1, 2, 3, 4, 5)
#> [1] 15
#> 
#> sum> ## In fact, you can pass vectors into several arguments, and everything gets added.
#> sum> sum(1:2, 3:5)
#> [1] 15
#> 
#> sum> ## If there are missing values, the sum is unknown, i.e., also missing, ....
#> sum> sum(1:5, NA)
#> [1] NA
#> 
#> sum> ## ... unless  we exclude missing values explicitly:
#> sum> sum(1:5, NA, na.rm = TRUE)
#> [1] 15

For the functions that comes from external packages, we first need to install the new packages. The most common way to do so is by executing install.packages("Package_Name"). Then when we want to load the functions, we start the script by executing library(Package_Name).

Finally, if we really do not find a suitable function in a package, we can create your functions following this general structure, but this will not be covered in this tutorial:

my_function <- function(<argument1>, <argument2>, ...){
  <here comes the definition of my function>
  return(<output of the definition>)
}

Exercises

• What is the function log() doing and from were does this function come from (base , other packages)? What is the function log() doing and from were does this function come from (base R, other packages)?

Solution

?log() #It takes the natural logarithm of the value, it comes from base R
log(10) 

• What are the mandatory arguments for the function plot()

Solution

?plot() # the coordinates points x and y are needed

• Is there help associated with the functions from a loaded package?

Hint

The function ggplot() comes from the package ggplot2

Solution

library(ggplot2)
?ggplot # Yes, there is also help for the imported functions

Optional exercises

• Create a function that print Hello World! when executing it

Solution

Hello <- function(){
  cat("Hello World!") # print("Hello World!") works too
}
Hello()
#> Hello World!

• Create a function that multiply the input by 4

Solution

multiplyeR <- function(x, y = 4){
  return(x * y)
}
multiplyeR(x = 2) # It works with values
#> [1] 8
multiplyeR(x = seq(1, 3, 1)) # But also vectors
#> [1]  4  8 12

Vectors

works with vector from which we can do our calculations. R works with vector from which we can do our calculations. Several ways exist to create a vector:

• Using c(), values are added next to each other and separated with a ,.

c(1, 2, 1, 4) # It works with integers (round numbers)
c(1.1, 2.4, 3.14652) # It works with floats (decimal numbers)
c("chocolate", "ice-cream") # It works with character
c(TRUE, FALSE) # It works with logical variables

• Using rep() to repeat the same values several times.

rep(x = 3, 2) # it reads: repeat 2 times the value x that is equal to 3
rep(x = "chocolate", 3) # it reads: repeat 3 times the value x that is equal to "chocolate"

• Using seq() to create a sequence of values. It only works for numeric values!

seq(from = 0, to = 10, by = 2) # it reads: create a sequence of values from 0 to 10 every 2 numbers
seq(from = -1, to = 1, by = 0.2) # it also works with negative values and decimal

• Combining all of the above

rep(x = c(seq(from = 2, to = 3, by = 0.2), 5), 2)
c(rep(x = "character", 5), "other character")
c(seq(from = 2, to = 10, by = 2), rep(x = 1000, 2), c(1, 4, 2))

Exercises

• Create a vector v1 that contains the values 1, 2, 3, 4, 6

Solution

v1 <- c(1, 2, 3, 4, 6)

• Create a vector v2 that contains 10 times the values 1, 2, 3, 4, 6

Solution

v2 <- rep(v1, 10)

• Create a vector v3 that repeats TRUE, FALSE 2 times

Solution

v3 <- rep(c(TRUE, FALSE), 2)

• Create a vector v4 that goes from 10 to 2000

Solution

v4 <- 10:2000
# or 
v4 <- seq(from  = 10, to = 2000, by = 1)

• Create a vector v5 that contains v1, v2, v3 and 2 times v4

Solution

v5 <- c(v1, v2, v3, rep(v4, 2))

Dataframe

Most likely, we will work with data stored in dataframes. A dataframe is composed of observations (rows) and variables (columns). We can see a dataframe like multiples vectors put togethers.

For example in the dataframe below (named df) is composed of 4 vectors:

1. Species that contains the species names
2. Abundance that contains the abundances of the species
3. Location that contains the location of the species
4. Date that contains the sampling date

#>         Species Abundance Location       Date
#> 1       Acartia        34     Askö 03-09-2024
#> 2 Pseudocalanus        12     Askö 04-09-2024
#> 3   Centropages        17     Askö 02-09-2024

We can access the individual columns (i.e., vectors) using $

df$Species
#> [1] "Acartia"       "Pseudocalanus" "Centropages"
df$Abundance
#> [1] 34 12 17
df$Location
#> [1] "Askö" "Askö" "Askö"
df$Date
#> [1] "03-09-2024" "04-09-2024" "02-09-2024"

Exercises

• Create a vector genus containing the character "Acartia", "Centropages", "Temora", "Acartia", "Centropages", "Temora"

Solution

genus = c("Acartia", "Centropages", "Temora", "Acartia", "Centropages", "Temora") 
# or genus = rep(c("Acartia", "Centropages", "Temora"), 2)

• Create a vector station containing the character "Askö", "Askö", "Askö", "Tjarnö", "Tjarnö", "Tjarnö"

Solution

station = c(rep("Askö",3),rep("Tjarnö", 3))

• Create a vector abundance containing the values 3, 10.2, 4, 2.3, 4, 9.4

Solution

abundance = c(3, 10.2, 4, 2.3, 4, 9.4)

• Combine all the vectors in a dataframe called df

Solution

df <- data.frame("Genus" = genus,
                 "Station" = station,
                 "Abundance" = abundance)

• Create a vector output that correspond to the column Abundance of the dataframe df. Is output similar to the vector abundance?

Solution

output <- df$Abundance # or df[[3]]
output == abundance
#> [1] TRUE TRUE TRUE TRUE TRUE TRUE

Importing data in Importing data in R

More often we enter our data in spreadsheets. We then need to import our data in R to process them.
To do so, we use the read.* function family.

A typical data import protocol looks like this:

1. Set the working directory with its absolute path

setwd("/Absolute/Path/To/Working/Directory")

2. Import your dataset in your environment

df <- read.csv("./Relative/Path/Dataset.csv")

3. Examine the structure of the data to see if the importation worked well

str(df)
head(df)
tail(df)

Exercises

• Import the dataset in your environment

Solution

df <- read.csv("./assets/zooplankton_seasonality.csv")

• How many rows and columns does this dataset contain?

Solution

The structure of the dataset shows that there is 7 variables (columns) and 2956 observations (rows)

str(df)
#> 'data.frame':    2956 obs. of  7 variables:
#>  $ Month_abb  : chr  "Jan" "Jan" "Jan" "Jan" ...
#>  $ Year       : int  2009 2009 2009 2009 2009 2009 2009 2009 2009 2009 ...
#>  $ Station    : chr  "BY15" "BY31" "BY5" "BY15" ...
#>  $ Coordinates: chr  "20.05000/57.33333" "18.23333/58.58812" "15.98333/55.25000" "20.05000/57.33333" ...
#>  $ Group      : chr  "Copepoda" "Copepoda" "Copepoda" "Copepoda" ...
#>  $ Taxa       : chr  "Acartia" "Acartia" "Acartia" "Centropages" ...
#>  $ Biomass    : num  6.65 1.82 5.56 5.74 1.23 ...

• What are the headers of the columns?

Solution

Both the stucture and the head show that the headers are: Month_abb, Year, Station, Coordinates, Group, Taxa, Biomass

head(df)
#>   Month_abb Year Station       Coordinates    Group        Taxa   Biomass
#> 1       Jan 2009    BY15 20.05000/57.33333 Copepoda     Acartia  6.650319
#> 2       Jan 2009    BY31 18.23333/58.58812 Copepoda     Acartia  1.816994
#> 3       Jan 2009     BY5 15.98333/55.25000 Copepoda     Acartia  5.562097
#> 4       Jan 2009    BY15 20.05000/57.33333 Copepoda Centropages  5.738561
#> 5       Jan 2009    BY31 18.23333/58.58812 Copepoda Centropages  1.228759
#> 6       Jan 2009     BY5 15.98333/55.25000 Copepoda Centropages 14.405224

• What is the last row?

Solution

To see the last row, use the tail function

tail(df)
#>      Month_abb Year Station       Coordinates     Group      Taxa    Biomass
#> 2951       Dec 2021    BY15 20.05000/57.33333  Copepoda    Temora 32.2266648
#> 2952       Dec 2021    BY31 18.23333/58.58812  Copepoda    Temora  7.6000062
#> 2953       Dec 2021     BY5 15.98333/55.25000  Copepoda    Temora 23.0666650
#> 2954       Dec 2021    BY15 20.05000/57.33333 Rotatoria Synchaeta  1.0400010
#> 2955       Dec 2021    BY31 18.23333/58.58812 Rotatoria Synchaeta  0.0800001
#> 2956       Dec 2021     BY5 15.98333/55.25000 Rotatoria Synchaeta  1.2900000